tính: (x+6) (x+6)=?
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\(x^2+x-2\)
\(=x^2-x+2x-2\)
\(=x\left(x-1\right)+2\left(x-1\right)\)
\(=\left(x-1\right)\left(x+2\right)\)
chì làm được 1 cách thôi
Theo bài ra ta có:\(M=\left(x+2\right)\left(3-x\right)\)
\(-x^2+x+6=4\\ \Leftrightarrow-x^2+x+2=0\\ \Leftrightarrow x^2-x-2=0\\ \Leftrightarrow\left(x+1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
Thay x=-1 vào M ta có:
\(M=\left(x+2\right)\left(3-x\right)=\left(1+2\right)\left(3-1\right)=3.2=6\)
Thay x=2 vào M ta có:
\(M=\left(x+2\right)\left(3-x\right)=\left(2+2\right)\left(3-2\right)=4.1=4\)
\(=\left(x^2-1^2\right)\left(x+2\right)=\left(x^2-1\right)\left(x+2\right)=x^3+2x^2-x+2\)
hằng đẳng thức số 3
\(x^2+x-6=x^2-2x+3x-6=x\left(x-2\right)+3\left(x-2\right)=\left(x-2\right)\left(x+3\right)\)
\(x^2+x-6=x^2+2.\frac{1}{2}x+\frac{1}{4}-\frac{25}{4}=\left(x+\frac{1}{2}\right)^2-\frac{25}{4}=\left(x+\frac{1}{2}-\frac{5}{2}\right)\left(x+\frac{1}{2}+\frac{5}{2}\right)=\left(x-2\right)\left(x+3\right)\)
x2+x-6=x3+3x-2x-6=x(x+3)-2(x+3)=(x+3)(x-2)
x2+x-6=\(x^2+x+\frac{1}{4}-6-\frac{1}{4}=\left(x+\frac{1}{2}\right)^2-\left(\frac{5}{2}\right)^2=\left(x+\frac{1}{2}-\frac{5}{2}\right)\left(x+\frac{1}{2}+\frac{5}{2}\right)=\left(x-2\right)\cdot\left(x+3\right)\)
a) \(\left(x-7\right)\left(x+2\right)\)
b) \(\left(2x-3\right)\left(x+2\right)\)
\(x^2+x-6\)
\(=x^2-2x+3x-6\)
\(=x\left(x-2\right)+3.\left(x-2\right)\)
\(=\left(x+3\right)\left(x-2\right)\)
Ta có: (x+6)(x+6)
\(=x^2+6x+6x+36\)
\(=x^2+12x+36\)